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Course in Probability Theory

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7 .2 LINDEBERG-FELLER THEOREM 1 223*7. F<strong>in</strong>d an example where L<strong>in</strong>deberg's condition is satisfied butLiapounov's is not for any 8 > 0 .In Exercises 8 to 10 below {X j , j > 1 } is a sequence of <strong>in</strong>dependent r .v .'s .8. For each j let X j have the uniform distribution <strong>in</strong> [-j, j] . Show thatL<strong>in</strong>deberg's condition is satisfied and state the result<strong>in</strong>g central limit theorem .9. Let X,. be def<strong>in</strong>ed as follows for some a > 1 :1fja , with probability 6 j2(a_1) each ;X~ - 10, with probability 1 - 3 j2(a-1)Prove that L<strong>in</strong>deberg's condition is satisfied if and only if a < 3/2 .* 10 . It is important to realize that the failure of L<strong>in</strong>deberg's conditionmeans only the failure of either (i) or (ii) <strong>in</strong> Theorem 7 .2 .1 with the specifiedconstants s, A central limit theorem may well hold with a different sequenceof constants . Letwith probability each ;12j 2with probability 12 each ;with probability 1 - 1 - 16 6j 2Prove that L<strong>in</strong>deberg's condition is not satisfied . Nonetheless if we take b, _11 3/18, then S„ /b„ converges <strong>in</strong> d ist . to (D . The po<strong>in</strong>t is that abnormally largevalues may not count! [HINT : Truncate out the abnormal value .]11 . Prove that fo x 2 d F (x) < oo implies the condition (11), but notvice versa .* 12 . The follow<strong>in</strong>g comb<strong>in</strong>atorial problem is similar to that of the numberof <strong>in</strong>versions . Let Q and '~P be as <strong>in</strong> the example <strong>in</strong> the text . It is standardknowledge that each permutation1a,2a2can be uniquely decomposed <strong>in</strong>to the product of cycles, as follows . Considerthe permutation as a mapp<strong>in</strong>g n from the set (1, . . ., n) onto itself suchthat 7r(j) = a j . Beg<strong>in</strong>n<strong>in</strong>g with 1 and apply<strong>in</strong>g the mapp<strong>in</strong>g successively,1 --* 7r (1) - 2r 2 (1) -* . . ., until the first k such that rr k (1) = 1 . Thus(1, 7r(l), 7r2(1), . . . , 7rk-1(1))

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